Mostrar que têm distribuição normal de inclinação
Seja e independentes. Mostre que têm uma distribuição normal de inclinação e encontre os parâmetros dessa distribuição.Y1∼SN(μ1,σ21,λ)Y1∼SN(μ1,σ12,λ)Y_1\sim SN(\mu_1,\sigma_1^2,\lambda)Y2∼N(μ2,σ22)Y2∼N(μ2,σ22)Y_2\sim N(\mu_2,\sigma_2^2)Y1+Y2Y1+Y2Y_1+Y_2 Como as variáveis aleatórias são independentes, tentei usar a convolução. SejaZ=Y1+Y2Z=Y1+Y2Z=Y_1+Y_2 fZ(z)=∫∞−∞2ϕ(y1|μ1,σ1)Φ(λ(y1−μ1σ1))ϕ(z−y1|μ2,σ22)dy1fZ(z)=∫−∞∞2ϕ(y1|μ1,σ1)Φ(λ(y1−μ1σ1))ϕ(z−y1|μ2,σ22)dy1f_Z(z)=\int_{-\infty}^{\infty}2\phi(y_1|\mu_1,\sigma_1)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\phi(z-y_1|\mu_2,\sigma_2^2)\,\text{d}y_1 Aqui e são os pdf e cdf normais padrão, respectivamente.ϕ()ϕ()\phi()Φ()Φ()\Phi() fZ(z)=∫∞−∞212πσ1−−−−√12πσ2−−−−√exp(−12σ21(y1−μ)2−12σ22((z−y1)2−μ)2)Φ(λ(y1−μ1σ1))dy1fZ(z)=∫−∞∞212πσ112πσ2exp(−12σ12(y1−μ)2−12σ22((z−y1)2−μ)2)Φ(λ(y1−μ1σ1))dy1f_Z(z)=\int_{-\infty}^{\infty}2\frac{1}{\sqrt{2\pi\sigma_1}}\frac{1}{\sqrt{2\pi\sigma_2}}exp\Big(-\frac{1}{2\sigma_1^2}(y_1-\mu)^2-\frac{1}{2\sigma_2^2}((z-y_1)^2-\mu)^2\Big)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1 Para notações simplificadas, deixek=212πσ1√12πσ2√k=212πσ112πσ2k=2\frac{1}{\sqrt{2\pi\sigma_1}}\frac{1}{\sqrt{2\pi\sigma_2}} fZ(z)=k∫∞−∞exp(−12σ21σ22(σ21(y1−μ1)2+σ22((z−y1)−μ2)2))Φ(λ(y1−μ1σ1))dy1=k∫∞−∞exp(−12σ21σ22(σ22(y21−2y1μ1+μ1)+σ21((z−y1)2−2(z−y1)μ2+μ22)))×Φ(λ(y1−μ1σ1))dy1=k∫∞−∞exp(−12σ21σ22(σ22(y21−2y1μ1+μ1)+σ21(z2−2zy1+y21−2zμ2+2y1μ2+μ22)))×Φ(λ(y1−μ1σ1))dy1fZ(z)=k∫−∞∞exp(−12σ12σ22(σ12(y1−μ1)2+σ22((z−y1)−μ2)2))Φ(λ(y1−μ1σ1))dy1=k∫−∞∞exp(−12σ12σ22(σ22(y12−2y1μ1+μ1)+σ12((z−y1)2−2(z−y1)μ2+μ22)))×Φ(λ(y1−μ1σ1))dy1=k∫−∞∞exp(−12σ12σ22(σ22(y12−2y1μ1+μ1)+σ12(z2−2zy1+y12−2zμ2+2y1μ2+μ22)))×Φ(λ(y1−μ1σ1))dy1\begin{align*}f_Z(z)&=k\int_{-\infty}^{\infty}\exp\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_1^2(y_1-\mu_1)^2+\sigma_2^2((z-y_1)-\mu_2)^2\Big)\Big)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1\\ &=k\int_{-\infty}^{\infty}\exp\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_2^2(y_1^2-2y_1\mu_1+\mu_1)+\sigma_1^2((z-y_1)^2-2(z-y_1)\mu_2+\mu_2^2)\Big)\Big)\\&\quad\times\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1=k\int_{-\infty}^{\infty} \exp\\&\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_2^2(y_1^2-2y_1\mu_1+\mu_1)+\sigma_1^2(z^2-2zy_1+y_1^2-2z\mu_2+2y_1\mu_2+\mu_2^2)\Big)\Big)\\&\quad\times\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1 \end{align*} …